Question

identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither. x = -3, x = -2, x = 0, x = 2, x = 3, x = 5, 5x / (x^3 + 5x^2 + 6x)

Explanation:

Step1: Factor the denominator

The denominator of the function $\frac{5x}{x^{3}+5x^{2}+6x}$ is $x^{3}+5x^{2}+6x = x(x^{2}+5x + 6)=x(x + 2)(x+3)$.

Step2: Find values that make denominator zero

Set the denominator equal to zero: $x(x + 2)(x + 3)=0$. The solutions are $x=0,x=-2,x=-3$. These are the potential points of discontinuity.

• For $x = 0$: Since $x$ is a factor in both the numerator and denominator, there is a hole at $x = 0$.
• For $x=-2$: Substituting $x=-2$ into the factored - form of the function, the denominator is zero and the numerator is non - zero. So, there is a vertical asymptote at $x=-2$.
• For $x=-3$: Substituting $x=-3$ into the factored - form of the function, the denominator is zero and the numerator is non - zero. So, there is a vertical asymptote at $x=-3$.
• For $x = 2$: Substitute $x = 2$ into the function $\frac{5x}{x^{3}+5x^{2}+6x}$, the denominator $2^{3}+5\times2^{2}+6\times2=8 + 20+12=40

eq0$, so $x = 2$ is neither a discontinuity.

• For $x = 3$: Substitute $x = 3$ into the function, the denominator $3^{3}+5\times3^{2}+6\times3=27+45 + 18=90

eq0$, so $x = 3$ is neither a discontinuity.

• For $x = 5$: Substitute $x = 5$ into the function, the denominator $5^{3}+5\times5^{2}+6\times5=125+125 + 30=280

eq0$, so $x = 5$ is neither a discontinuity.

Answer:

$x = 5$: neither
$x = 3$: neither
$x = 2$: neither
$x = 0$: hole
$x=-2$: asymptote
$x=-3$: asymptote